The NLP Procedure

Example 6.1 Using the DATA= Option

This example illustrates the use of the DATA= option. The Bard function (refer to Moré, Garbow, and Hillstrom (1981)) is a least squares problem with $n=3$ parameters and $m=15$ functions $f_ k$ :

$f(x) = \frac{1}{2} \sum _{k=1}^{15} f_ k^2(x) , \quad x = (x_1,x_2,x_3)$

where

$f_ k(x) = y_ k - \left( x_1 + \frac{u_ k}{v_ k x_2 + w_ k x_3} \right)$

with $u_ k=k$ , $v_ k=16-k$ , $w_ k=\min (u_ k, v_ k)$ , and

$y= ( .14, .18, .22, .25, .29, .32, .35, .39, .37, .58, .73, .96, 1.34, 2.10, 4.39 )$

The minimum function value $f(x^*) =$ 4.107E–3 is at the point $(0.08,1.13,2.34)$ . The starting point $x^0 = (1,1,1)$ is used. The following is the naive way of specifying the objective function.

proc nlp tech=levmar;
   lsq y1-y15;
   parms x1-x3 = 1;
   tmp1 = 15 * x2 + min(1,15) * x3;
   y1 = 0.14 - (x1 + 1 / tmp1);
   tmp1 = 14 * x2 + min(2,14) * x3;
   y2 = 0.18 - (x1 + 2 / tmp1);
   tmp1 = 13 * x2 + min(3,13) * x3;
   y3 = 0.22 - (x1 + 3 / tmp1);
   tmp1 = 12 * x2 + min(4,12) * x3;
   y4 = 0.25 - (x1 + 4 / tmp1);
   tmp1 = 11 * x2 + min(5,11) * x3;
   y5 = 0.29 - (x1 + 5 / tmp1);
   tmp1 = 10 * x2 + min(6,10) * x3;
   y6 = 0.32 - (x1 + 6 / tmp1);
   tmp1 = 9 * x2 + min(7,9) * x3;
   y7 = 0.35 - (x1 + 7 / tmp1);
   tmp1 = 8 * x2 + min(8,8) * x3;
   y8 = 0.39 - (x1 + 8 / tmp1);
   tmp1 = 7 * x2 + min(9,7) * x3;
   y9 = 0.37 - (x1 + 9 / tmp1);
   tmp1 = 6 * x2 + min(10,6) * x3;
   y10 = 0.58 - (x1 + 10 / tmp1);
   tmp1 = 5 * x2 + min(11,5) * x3;
   y11 = 0.73 - (x1 + 11 / tmp1);
   tmp1 = 4 * x2 + min(12,4) * x3;
   y12 = 0.96 - (x1 + 12 / tmp1);
   tmp1 = 3 * x2 + min(13,3) * x3;
   y13 = 1.34 - (x1 + 13 / tmp1);
   tmp1 = 2 * x2 + min(14,2) * x3;
   y14 = 2.10 - (x1 + 14 / tmp1);
   tmp1 = 1 * x2 + min(15,1) * x3;
   y15 = 4.39 - (x1 + 15 / tmp1);
run;

A more economical way to program this problem uses the DATA= option to input the 15 terms in $f(x)$ .

data bard;
   input r @@;
      w1 = 16. - _n_;
      w2 = min(_n_ , 16. - _n_);
      datalines;
.14  .18  .22  .25  .29  .32  .35  .39
.37  .58  .73  .96 1.34 2.10 4.39
;

proc nlp data=bard tech=levmar;
   lsq y;
   parms x1-x3 = 1.;
   y = r - (x1 + _obs_ / (w1 * x2 + w2 * x3));
run;

Another way you can specify the objective function uses the ARRAY statement and an explicit do loop, as in the following code.

proc nlp tech=levmar;
   array r[15] .14  .18  .22  .25  .29  .32  .35  .39  .37  .58
               .73  .96 1.34 2.10 4.39 ;
   array y[15] y1-y15;
   lsq y1-y15;
   parms x1-x3 = 1.;
   do i = 1 to 15;
      w1 = 16. - i;
      w2 = min(i , w1);
      w3 = w1 * x2 + w2 * x3;
      y[i] = r[i] - (x1 + i / w3);
   end;
run;